A m = 70.5 kg object is released from rest at a distance h = 0.813256 R above th

Question

A m = 70.5 kg object is released from rest at
a distance h = 0.813256 R above the Earth’s
surface.
The acceleration of gravity is 9.8 m/s2 .
For the Earth, RE = 6.38 × 106 m, M =
5.98 × 1024 kg. The gravitational acceleration
at the surface of the earth is g = 9.8 m/s2.
Find the speed of the object when it strikes
the Earth’s surface. Neglect any atmospheric
friction.
Caution: You must take into account that
the gravitational acceleration depends on
distance between the object and the center of the
earth.
Answer in units of m/s.

I get 5542m/s as my answer and the system does not accept it. Can someone help me get the right answer?

Explanation / Answer

Gravitational potential energy is -GMm/r

The small g = 9.81 is irrelelvant to this problem

so at the high point, U = -(6.6738 * 10-11 m3/kg s2)(5.98 * 1024 kg)(70.5 kg)/(1.813256 * 6.38 * 106 m)

= -2.4321 * 109 J

At the low point, do the same calculation only the denominator doesn't have the 1.813256 factor, you get

-4.4100 * 109 J

So the object picks up 1.9779 * 109 J of kinetic energy. That is equal to (1/2)mv2. Solving for v, with m = 70.5 kg, I get 7491 m/s.

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